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We will learn how to find the intercepts on the axes made by a circle.
The lengths of intercepts made by the circle x2 + y2 + 2gx + 2fy + c = 0 with X and Y axes are 2βg2βc and 2βf2βc respectively.
Proof:
Let the given equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 β¦β¦β¦. (1)
Clearly, the centre of the circle is c (-g, -f) and the radius = βg2+f2βc
Let AB be the intercept made by the given circle on x-axe.
Since on x-axis, y = 0. Therefore, x-coordinates of the points A and B are the
roots of the equation x2 + 2gx + c = 0.
Let x1 and x2 be the x-coordinates of the points A and B respectively. Then, x1 and x2 also the roots of the equation x2 + 2gx + c = 0.
Therefore, x1 + x2 = - 2g and x1x2 = c
Clearly the intercept on x-axis = AB
= x2 - x1 = β(x2βx1)2
= β(x2+x1)2β4x1x2
= β4g2β4c
= 2βg2βc
Therefore, the intercept made by the circle (1) on the x-axis = 2βg2βc
Again,
Let DE be the intercept made by the given circle on y-axe. Since on y-axis, x = 0. Therefore, y-coordinates of the points D and E are the roots of the equation y2 + 2fy + c = 0.
Let y1 and y2 be the x-coordinates of the points D and E respectively. Then, y1 and y2 also the roots of the equation y2 + 2fy + c = 0
Therefore, y1 + y2 = - 2f and y1y2 = c
Clearly the intercept on y-axis = DE
= y2 - y1 = β(y2βy1)2
= β(y2+y1)2β4y1y2
= β4f2β4c
= 2βf2βc
Therefore, the intercept made by the circle (1) on the y-axis = 2βf2βc
Solved examples to find the intercepts made by a given circle on the co-ordinate axes:
1. Find the length of the x-intercept and y-intercept made by the circle x2 + y2 - 4x -6y - 5 = 0 with the co-ordinate axes.
Solution:
Given equation of the circle is x2 + y2 - 4x -6y - 5 = 0.
Now comparing the given equation with the general equation of the circle x2 + y2 + 2gx + 2fy + c = 0, we get g = -2 and f = -3 and c = -5
Therefore, length of the x-intercept = 2βg2βc = 2β4β(β5) = 2β9 = 6.
The length of the y-intercept = 2βf2βc = 2β9β(β5) = 2β14.
2. Find the equation of a circle which touches the y-axis at a distance -3 from the origin and cuts an intercept of 8 units with the positive direction of x-axis.
Solution:
Let the equation of the circle be x2 + y2 + 2gx + 2fy + c = 0 β¦β¦β¦β¦β¦.. (i)
According to the problem, the equation (i) touches the y-axis
Therefore, c = f2 β¦β¦β¦β¦β¦β¦β¦ (ii)
Again, the point (0, -3) lies on the circle (i).
Therefore, putting the value of x = 0 and y = -3 in (i) we get,
9 - 6f + c = 0 β¦β¦β¦β¦β¦β¦β¦β¦ (iii)
From (ii) and (iii), we get 9 - 6f + f2 = 0 β (f - 3)2 = 0 β f - 3 = 0 β f = 3
Now putting f = 3 in (i) we get, c = 9
Again, according to the problem the equation of the circle (i) cuts an intercept of 8 units with the positive direction of x-axis.
Therefore,
2βg2βc = 8
β 2βg2β9 = 8
β βg2β9 = 4
β g2 - 9 = 16, [Squaring both sides]
β g2 = 16 + 9
β g2 = 25
β g = Β±5.
Hence, the required equation of the circle is x^2 + y^2 Β± 10x + 6y + 9 = 0.
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